The point on the curve $x^2=8 y$ which is nearest to the point (2, 4) is:

(4, 2)

The correct answer is Option (3) → (4, 2)

$x^2 = 8y \Rightarrow y = \frac{x^2}{8}$

$D^2 = (x-2)^2 + \left(\frac{x^2}{8} - 4\right)^2$

$\frac{d}{dx}(D^2) = 2(x-2) + 2\left(\frac{x^2}{8} - 4\right)\cdot \frac{x}{4}$

$= 2(x-2) + \frac{x}{2}\left(\frac{x^2}{8} - 4\right)$

$= 2x - 4 + \frac{x^3}{16} - 2x = \frac{x^3}{16} - 4$

$\frac{x^3}{16} - 4 = 0 \Rightarrow x^3 = 64 \Rightarrow x = 4$

$y = \frac{4^2}{8} = 2$

$\text{Nearest point is } (4,2)$